RunTime Complexity

Hi there Buddies.
I've already answered the first 2 Functions But I wanna know if my answers are true
But I couldnt find a solution for the 3rd function
Here is the Question
1. Let n=2k , Determine RunTime Complexity for each of the following 3 functions as a function of k.
2. Calculate  prod as function of k in each case.

static int F1(int n)
            {
                  int prod=1;
                  while (n>1)
                  {
                        prod = prod *n;
                        n=n/2;
                  }
                  return prod;
            }

static int F2(int n)
            {
                  int prod=1; int m;
                  while (n>1)
                  {
                        m=n;
                        while (m>1)
                        {
                              prod = prod *2;
                              m=m/2;
                        }
                        n=n/2;
                  }
                  return prod;
            }

static double F3(int n)
            {
                  double prod=1; int x=0;
                  for (int i=1; i<=n; i++)
                        for (int j=1; j<=n*n; j++)
                              if ((j % Math.Sqrt(n))==0)
                                    for (int m=1; m<=j; m++)
                                          prod=prod *2;
                  return prod;
            }



My answers for the first 2 Functions

First Function :
Complexity as a function of n is log2(n)
Complexity as a function of k is log2(2k) = k   ,,, so It’s O(k)
Prod = 2k * 2k-1 * 2k-2 * 2k-3 * 2k-4 *2k-5 *………* 21  = 2k  -  2
Second Function :
Complexity as a function of n  is n*log(n)
Complexity as a function of 2k is 2k*log(2k)  = 2k * k  ,,, so It’s O(2k)
Prod =  2k * 2k-1 * 2k-2 * 2k-3 * 2k-4 *2k-5 *………* 21   = 2k  -  2


Thanks
LVL 2
nikorbaAsked:
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ozoCommented:
I gather that 2k means 2^k or 2**k

> Second Function :
> Complexity as a function of n  is n*log(n)
I don't think that's right
0
ozoCommented:
> if ((j % Math.Sqrt(n))==0)
How many times would this be true?
0
ozoCommented:
> Prod = 2k * 2k-1 * 2k-2 * 2k-3 * 2k-4 *2k-5 *………* 21  = 2k  -  2
Are you saying Prod is less than 2k?
0
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Sinoj SebastianCTO & OpenERP Project managerCommented:
Sorry, not an answer, just a point

log2(2k) = 1+log2(k)  {not k; log2(2^k)=k}
0
nikorbaAuthor Commented:
Hi there ,,, well , there are some mistakes in the Question ,,, because I copied it from the word Document ,,, so I thought that the exponent will appear correctly ...I will fix it now
Here is the Question
1. Let n=2^k , Determine RunTime Complexity for each of the following 3 functions as a function of k.
2. Calculate  prod as function of k in each case.

static int F1(int n)
            {
                  int prod=1;
                  while (n>1)
                  {
                        prod = prod *n;
                        n=n/2;
                  }
                  return prod;
            }

static int F2(int n)
            {
                  int prod=1; int m;
                  while (n>1)
                  {
                        m=n;
                        while (m>1)
                        {
                              prod = prod *2;
                              m=m/2;
                        }
                        n=n/2;
                  }
                  return prod;
            }

static double F3(int n)
            {
                  double prod=1; int x=0;
                  for (int i=1; i<=n; i++)
                        for (int j=1; j<=n*n; j++)
                              if ((j % Math.Sqrt(n))==0)
                                    for (int m=1; m<=j; m++)
                                          prod=prod *2;
                  return prod;
            }



My answers for the first 2 Functions

First Function :
Complexity as a function of n is log2(n)
Complexity as a function of k is log2(2^k) = k   ,,, so It’s O(k)
Prod = 2^k * 2^(k-1) * 2^(k-2) * 2^(k-3 )* 2^(k-4) *2^(k-5 )*………* 2^1  = 2^(k)  -  2
Second Function :
Complexity as a function of n  is n*log(n)
Complexity as a function of k is 2^k*log(2^k)  = (2^k )* k  ,,, so It’s O(2^k)
Prod = 2^k * 2^(k-1) * 2^(k-2) * 2^(k-3 )* 2^(k-4) *2^(k-5 )*………* 2^1  = 2^(k)  -  2

Thanks
0
Infinity08Commented:
>> Prod = 2^k * 2^(k-1) * 2^(k-2) * 2^(k-3 )* 2^(k-4) *2^(k-5 )*………* 2^1  = 2^(k)  -  2

Not really (as ozo already noted).

           2^(k(k + 1)/2)

is more like it ... can you explain why ?


>> Complexity as a function of n  is n*log(n)

That's not correct ... you have nested loops of the same type as in the first function ... what does that give complexity wise ?
0

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nikorbaAuthor Commented:
OZo didn't saw my 2nd comment :)
I think the 1st two functions are correct
0
Infinity08Commented:
>> I think the 1st two functions are correct

No, see my previous post ...
0
nikorbaAuthor Commented:
>>>No, see my previous post ...

I saw it,,,
try to solve it numerically
try a number with the F1
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Infinity08Commented:
>> try to solve it numerically
>> try a number with the F1

What do you mean ?

You said :

      Prod = 2^k * 2^(k-1) * 2^(k-2) * 2^(k-3 )* 2^(k-4) *2^(k-5 )*………* 2^1  = 2^(k)  -  2

And that's simply not true ... (see my earlier post)
0
ozoCommented:
there were answers
0
Infinity08Commented:
I agree ... nikorba apparently didn't believe us when we pointed out his mistakes ...
0
nikorbaAuthor Commented:
I still think that my answers were right ,,, that;s it
0
Infinity08Commented:
Then I wonder why you were asking our help, if you can't accept what we have to say ?
0
nikorbaAuthor Commented:
I appreciate ur help ,,, at first I wasnt sure about my answers so I asked u
but after a while the I asked the lecturer and he said my answers were right
I really thank u so much for ur help but that what happened
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Infinity08Commented:
If your teacher says that this :

    Prod = 2^k * 2^(k-1) * 2^(k-2) * 2^(k-3 )* 2^(k-4) *2^(k-5 )*………* 2^1  = 2^(k)  -  2

is correct, then he's wrong :)

2^(k)  -  2 is smaller than 2^(k), while Prod is a lot larger. How can those two be equal ?

Try filling in k=3 for example. You say that :

    Prod = 2^3 * 2^(2) * 2^(1)  = 2^(3)  -  2
    Prod = 2^6  = 2^3  -  2
    Prod = 64  = 6

How can 64 be equal to 6 ???
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VenabiliCommented:
OK - this is one of these "why the hell I recommended delete" :)

So what we do with the question?
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Infinity08Commented:
If nikorba still wants our help, then I'll be glad to provide that ... if not, then this question can be PAQ'd for ozo as he was first :)
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nikorbaAuthor Commented:
I'm sorry I discovered that I was wrong
sorry everyone :)
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